How do you list all possible roots and find all factors and zeroes of #5x^3+29x^2+19x-5#?

1 Answer
Jun 23, 2016

Answer:

#5x^3+29x^2+19x-5 = (x+1)(5x-1)(x+5)#

with zeros #x=-1#, #x=1/5# and #x=-5#

Explanation:

#f(x) = 5x^3+29x^2+19x-5#

Note that if you reverse the signs on the terms of odd degree then the sum of the coefficients is #0#.

That is: #-5+29-19-5 = 0#

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Hence #f(-1) = 0# and #(x+1)# is a factor:

#5x^3+29x^2+19x-5 = (x+1)(5x^2+24x-5)#

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To factor the remaining quadratic use an AC method:

Look for a pair of factors of #AC = 5*5 = 25# which differ by #B=24#.

The pair #25, 1# works, in that #25xx1=25# and #25-1 = 24#.

Use this pair to split the middle term and factor by grouping:

#5x^2+24x-5#

#=5x^2+25x-x-5#

#=(5x^2+25x)-(x+5)#

#=5x(x+5)-1(x+5)#

#=(5x-1)(x+5)#

Hence zeros #x=1/5# and #x = -5#